EXAM FM SAMPLE SOLUTIONS - Society of Actuaries

SOCIETY OF ACTUARIES EXAM FM FINANCIAL MATHEMATICS

EXAM FM SAMPLE SOLUTIONS This set of sample questions includes those published on the interest theory topic for use with previous versions of this examination. Questions from previous versions of this document that are not relevant for the syllabus effective with the October 2022 administration have been deleted. The questions have been renumbered. Some of the questions in this study note are taken from past SOA examinations. These questions are representative of the types of questions that might be asked of candidates sitting for the Financial Mathematics (FM) Exam. These questions are intended to represent the depth of understanding required of candidates. The distribution of questions by topic is not intended to represent the distribution of questions on future exams. The following model solutions are presented for educational purposes. Alternative methods of solution are acceptable. In these solutions, sm is the m-year spot rate and m ft is the m-year forward rate, deferred t years. Update history: October 2022: Questions 208-275 were added January 2023: Question 204 was deleted June 2023 Questions 276-385 were added

Copyright 2023 by the Society of Actuaries.

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1. Solution: C Given the same principal invested for the same period of time yields the same accumulated

value, the two measures of interest i(2) = 0.04 and must be equivalent, which means:

1+

i(2) 2

2

= e over

a

one-year

period.

Thus,

e

= 1 +

i

(2)

2

2

= 1.022

= 1.0404

= ln= (1.0404) 0.0396.

2. Solution: E

From basic principles, the accumulated values after 20 and 40 years are

100[(1

+

i)20

+

(1

+

i)16

+

+

(1

+

i)4

]

=100

(1

+ i)4 - (1+ i)24 1- (1+ i)4

100[(1

+

i)40

+

(1

+

i)36

+

+

(1 +

i)4

]

=100

(1

+ i)4 - (1+ i)44 1- (1+ i)4

.

The ratio is 5, and thus (setting x= (1+ i)4 )

=5

(= 1+ i)4 - (1+ i)44 (1+ i)4 - (1+ i)24

x - x11 x - x6

5x - 5x6 =x - x11

5 - 5x5 =1- x10

x10 - 5x5 + 4 =0

(x5 -1)(x5 - 4) = 0.

Only the second root gives a positive solution. Thus x5 = 4

x = 1.31951

X 1= 00 1.31951-1.3195111 6195. 1 - 1.31951

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Annuity symbols can also be used. Using the annual interest rate, the equation is

100 s40 = 5(100) s20

a4

a4

(1+ i)40 -1 = 5 (1+ i)20 -1

i

i

(1+ i)40 - 5(1+ i)20 + 4 = 0

(1+ i)20 = 4

and the solution proceeds as above.

3. Solution: C

Eric's

(compound)

interest

in

the

last

6

months

of

the

8th

year

is

1001 +

i 2

15

i 2

.

Mike's (simple) interest for the same period is 200 i . 2

Thus,

100 1 +

i 2

15

i 2

= 200 i 2

1 +

i 2

15

= 2

1+ i =1.047294 2

=i 0= .09459 9.46%.

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4. Solution: C

= 77.1

v

(

Ia

) n

+

nvn+1 i

=

v

a n

- nvn

+

nvn+1

i i

=an - nvn+1 + nvn+1

ii

i

= a=n 1- v=n 1- vn i i2 0.011025

0.85003= 1- vn

1.105-n = 0.14997

n = - ln(0.14997) = 19. ln(1.105)

To obtain the present value without remembering the formula for an increasing annuity, consider the payments as a perpetuity of 1 starting at time 2, a perpetuity of 1 starting at time 3, up to a perpetuity of 1 starting at time n + 1. The present value one period before the start of each perpetuity is 1/i. The total present value is (1/ i)(v + v2 + + vn ) =(1/ i)an .

5. Solution: C

The interest earned is a decreasing annuity of 6, 5.4, etc. Combined with the annual deposits of

100, the accumulated value in fund Y is

6(Ds)10 0.09 +100s10 0.09

6

10(1.09)10 -

0.09

s10

0.09

+ 100 (15.19293)

= 565.38 +1519.29

= 2084.67.

6. Solution: D For the first 10 years, each payment equals 150% of interest due. The lender charges 10%, therefore 5% of the principal outstanding will be used to reduce the principal.

At the end of 10 years, the amount outstanding is 1000(1- 0.05)10 = 598.74 .

Thus, the equation of value for the last 10 years using a comparison date of the end of year 10 is 5= 98.74 X= a10 10% 6.1446 X

X = 97.44.

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7. Solution: B The book value at time 6 is the present value of future payments: BV6 = 10, 000v4 + 800a4 0.06 = 7920.94 + 2772.08 = 10, 693.

The interest portion is 10,693(0.06) = 641.58.

8. Solution: A

The value of the perpetuity after the fifth payment is 100/0.08 = 1250. The equation to solve is:

1250=

X

(v

+

1.08v2

+

+

1.0824

v25

)

=

X

(v

+

v

+

+

v=)

X (25) /1.08

=X 5= 0(1.08) 54.

9. Solution: C Equation of value at end of 30 years: 10(1- d / 4)-40 (1.03)40 + 20(1.03)30 = 100

10(1- d / 4)-40 = [100 - 20(1.03)30 ] / 1.0340 = 15.7738

1-= d / 4 1.5773= 8-1/40 0.98867

d = 4(1- 0.98867) = 0.0453 = 4.53%.

10. Solution: E

The accumulati= on function is a(t) e= xp 0t (s2 /100)ds exp(t3 / 300).

The accumulated value of 100 at time 3 is 100 exp(33 / 300) = 109.41743. The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus

(109.41743 + X )[a(6) / a(3) -1] =X

(109.41743 + X )(2.0544332 /1.0941743 -1) =X

(109.41743 + X )0.877613 = X

96.026159 = 0.122387 X

X = 784.61.

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